Consider the TRS R consisting of the rewrite rules

1: f(x,y,w,w,a) -> g1(x,x,y,w)
2: f(x,y,w,a,a) -> g1(y,x,x,w)
3: f(x,y,a,a,w) -> g2(x,y,y,w)
4: f(x,y,a,w,w) -> g2(y,y,x,w)
5: g1(x,x,y,a) -> h(x,y)
6: g1(y,x,x,a) -> h(x,y)
7: g2(x,y,y,a) -> h(x,y)
8: g2(y,y,x,a) -> h(x,y)
9: h(x,x) -> x

There are 8 dependency pairs:

10: F(x,y,w,w,a) -> G1(x,x,y,w)
11: F(x,y,w,a,a) -> G1(y,x,x,w)
12: F(x,y,a,a,w) -> G2(x,y,y,w)
13: F(x,y,a,w,w) -> G2(y,y,x,w)
14: G1(x,x,y,a) -> H(x,y)
15: G1(y,x,x,a) -> H(x,y)
16: G2(x,y,y,a) -> H(x,y)
17: G2(y,y,x,a) -> H(x,y)

The approximated dependency graph contains no SCCs
and hence the TRS is trivially terminating.